What I meant by "effectively" is this:
Let us say for the moment that we have a way of measuring, expressing, and replicating the amount of drag created by an inoperative engine. Let's call that drag 'D'.
With both engines operating normally and producing 100% thrust, you get a climb rate of XXX fpm at Vyse (just to use a reference V speed). For these purposes, the exact value of XXX does not matter.
If, while climbing at 100% thrust on both engines, you somehow toss out and additional D worth of drag, you will no longer be able to maintain XXX fpm at Vyse. The thrust required to maintain XXX fpm at Vyse will be something in excess of 100% of what you have. For the sake of argument, let's say that a D increase in drag requires 130% of the thrust you have available to climb at XXX fpm at Vyse.
Now, instead of just tossing D drag out in the slipstream, you also reduce the available thrust by 50% (you just lost an engine). Because of the D increase in drag, you still need 130% of the thrust you normally have available, but you only have 50% of normal and less than 39% of the thrust required to continue to climb at XXX fpm at Vyse with the additional D increase in drag.
In summary:
Without D, you need 100% trust for XXX fpm at Vyse
With D you need 130% thrust for XXX fpm at Vyse
With one engine failed, you have only 38.5% of thrust required for XXX fpm at Vyse, effectively (not literally) reducing your thrust to something significantly less than 50%.